Question: Let $f$ be a function defined for all real numbers except for $4$. Also let $f'$, the derivative of $f$, be defined as $f'(x)=\dfrac{(x+1)^2}{x-4}$. On which intervals is $f$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $(-1,4)$ only (Choice B) B $(-\infty,-1)$ only (Choice C) C $(-\infty,-1)$ and $(4,\infty)$ (Choice D) D $(4,\infty)$ only (Choice E) E The entire domain of $f$
Answer: We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $f'(x)=\dfrac{(x+1)^2}{x-4}$ and that $f$ is undefined at $x=4$. $f'(x)=0$ for $x=-1$. Our critical point is $x=-1$, and we should also consider $x=4$. Our points divide the number line into three intervals: $\llap{-}6$ $\llap{-}4$ $\llap{-}2$ $0$ $2$ $4$ $6$ $(-\infty, \llap{-}1)$ $( \llap{-}1,4)$ $(4,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,-1)$ $x=-2$ $f'(-2)=-\dfrac{1}{6}<0$ $f$ is decreasing. $\searrow$ $(-1,4)$ $x=0$ $f'(0)=-\dfrac{1}{4}<0$ $f$ is decreasing. $\searrow$ $(4,\infty)$ $x=5$ $f'(5)=36>0$ $f$ is increasing. $\nearrow$ In conclusion, $f$ is increasing over the interval $(4,\infty)$ only.